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Recently as an interview question for an RF engineer job, I was asked the following question.

What are the dangers of having a local oscillator power level much higher than that of the RF power level during mixing?

I did not feel as though I had a great answer to this question. The main concerns I could think of are as follows and are based on Poser's Microwave Circuits equation for the output of a single-ended Schottky diode mixer; (using the small signal approximation) (13.100)

\$ i(t) = I_0 + G_d[v_{RF}(t)+v_{LO}(t) +\dfrac{G'_d}{2}[v_{RF}(t)+v_{LO}(t)]^2 + ... \$

\$ i(t) = I_0 + G_d[v_{RF}(t)+v_{LO}(t) + \dfrac{G'_d}{4}(V_{RF}^2(1+cos(2w_{RF}t))+V_{LO}^2(1+cos(2w_{LO}t)) + 2V_{RF}V_{LO}cos((w_{RF}-w_{LO})t) + 2V_{RF}V_{LO}cos((w_{RF}+w_{LO})t) \$

I understand for a balanced mixer some of these terms could be removed/problems alleviated.

  1. Coupling from LO to RF input could lead to strong radiation of LO signal
  2. At the output of the mixer, there will be a term that is a copy of the LO signal. This term should be filtered out, but maybe if it's strong enough could still show up at the ADC
  3. There may also be terms that are the square voltage of the LO signal and with either 0 frequency (DC) or double the frequency of the LO. These terms also should be filtered out, but if strong enough, perhaps these components could also still overpower the RF signal after filtering. In the case of the DC component, it could also bias the diodes/transistors used
  4. In the case of unexpected components showing up in the signal, it could be possible for the ADC to be saturated/a loss in dynamic range

If anyone can think of additional possible answers or explain why mine are wrong, please do so.

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  • \$\begingroup\$ Did you start scribbling those equations in the job interview? (I'm trying to understand the relevance of you mentioning the job interview if any). In short, what was your job-interview reply (you have tagged it). \$\endgroup\$
    – Andy aka
    Jan 23, 2023 at 10:53

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In a conventional schottky diode mixer, you always have the LO at a much higher power than the RF signal. Or to put it another way, you run the diodes at their optimum power rating, and choose an RF signal level that's well below that. For a +13dBm LO mixer, -10dBm would be quite a high RF level, and you would usually run lower than that.

The main problem of a high LO level (I wouldn't really use the word danger, it's a feature that you have to design around) is LO leakage to the other ports. Downstream leakage can overload LNAs following the mixer. You deal with this by balancing the mixer, and filtering. Ideally, the filter should not present a bad VSWR at any frequency to the mixer, and this is where the filter designers earn their money.

I once built a 0-4 GHz mixer that had PIN and varactor balancing components within the mixer itself, to actively balance the +17dBm LO to an IF leakage of better than -40dBm, in a spectrum analyser application. It was followed by a directional filter, an unusual filter design that uses two λ/4 couplers as part of a resonator ring, to extract an IF, while presenting a reasonable match at all frequencies to the mixer.

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  • \$\begingroup\$ Hi @Neil_UK thank you for the answer, if you don't mind me asking how were you able to use a \$ \dfrac{\lambda}{4} \$ coupler in such a wide band filter? \$\endgroup\$
    – CMH12
    Jan 25, 2023 at 12:52
  • \$\begingroup\$ @CMH12 It was an upconverter, input DC-4 GHz, LO 5 GHz to 9 GHz, IF out 5 GHz with 100 MHz bandwidth, so no problem with coupler bandwidth at all. The main arm of the coupler went to a good load, and the ring only resonated to pick off the narrow IF part of the spectrum - so good match at all frequencies DC to light. The problem was the input really did go down to DC, so at 5 GHz, the LO leaked straight into the IF. \$\endgroup\$
    – Neil_UK
    Jan 25, 2023 at 13:13

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